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Bipartite graph $G=(A,B)$ with $delta(A)=3n/2$ and no $C_4$ has a matching which saturate each vertex in $A$.

Show that a finite regular bipartite graph has a perfect matchingShowing a bipartite graph has a perfect matching.bipartite graph has perfect matchingEdge and vertex connectivity of bipartite graphMatching in a bipartite graphMatching in bipartite graphProof bipartite graph matchingIf graph G is bipartite with largest vertex degree $Delta$(G) , then $chi'(G)$ = $Delta$(G)Maximum matching for bipartite graphHow would we prove that the following bipartite graph has a perfect matching?

3

2

$begingroup$

Say $G$ is a bipartite graph with bipartition $(A,B)$ and $G$ is $C_4$-free. Prove that if every vertex in $A$ has degree at least $frac32 n$ and $|A|leq n^2$, then $G$ has a matching which uses every vertex in $A$.

---

My proof:

Use the Hall marriage theorem. Take any $Xsubseteq A$ and let $|X|=k$. Also let $Y=N(X)$ and $|Y|=l$. Let us prove that $lgeq k$. Assume that $l<k$.

Let $Y^*$ be a set of all unordered pairs $y_i,y_j$, $ine j$ of elements in $Y$, and connect a pair $y_i,y_j$ with $xin X$ iff both $y_i$ and $y_j$ are adjacent with $x$ in $G$. Then the degree of each pair in this new bipartite graph $G^*$ (on vertex set $X cup Y^*$) is at most $1$ (since there is no $C_4$ in $G$) and the degree of each $xin X$ is at least $displaystyle3nover 2choose 2$. So we have $$kcdot 3nover 2choose 2leq lchoose 2 .$$
Since we assume $l<k$ we have $$3nover 2choose 2< k-1over 2$$ so $$3n(3n-2)over 4 < k-1 .$$ Since $kleq n^2$ we have $$3n(3n-2) leq 4n^2-8$$so
$$5n^2leq 6n-8$$ which is obviously not true.

Edit: After Darij's confirmation that the proof is correct, I'm will award any solution with better bound than $delta (G)=nsqrt2$ (instead of $3n/2$).

combinatorics discrete-mathematics proof-verification graph-theory

share|cite|improve this question

edited 12 hours ago

Maria Mazur

asked May 25 '18 at 15:10

Maria MazurMaria Mazur

46.7k1260120

$endgroup$

This question has an open bounty worth +100
reputation from Maria Mazur ending ending at 2019-03-17 18:50:50Z">in 4 days.

Looking for an answer drawing from credible and/or official sources.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  How exactly do you prove this for $n$ instead of $3n/2$ ? I don't get the inequalities to product a contradiction.
  $endgroup$
  – darij grinberg
  2 days ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  We are talking now about $delta = 3n/2$ or $n$?
  $endgroup$
  – Maria Mazur
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I am talking about $n$.
  $endgroup$
  – darij grinberg
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes, I see. My bad. Anyway, is the rest correct?
  $endgroup$
  – Maria Mazur
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The rest is correct. You may want to compute the optimal bound for your proof to work; I'm pretty sure it falls closer to $n$ than to $3n/2$.
  $endgroup$
  – darij grinberg
  2 days ago
  

  
  
  
  

 | 
show 2 more comments

3

2

$begingroup$

Say $G$ is a bipartite graph with bipartition $(A,B)$ and $G$ is $C_4$-free. Prove that if every vertex in $A$ has degree at least $frac32 n$ and $|A|leq n^2$, then $G$ has a matching which uses every vertex in $A$.

---

My proof:

Use the Hall marriage theorem. Take any $Xsubseteq A$ and let $|X|=k$. Also let $Y=N(X)$ and $|Y|=l$. Let us prove that $lgeq k$. Assume that $l<k$.

Let $Y^*$ be a set of all unordered pairs $y_i,y_j$, $ine j$ of elements in $Y$, and connect a pair $y_i,y_j$ with $xin X$ iff both $y_i$ and $y_j$ are adjacent with $x$ in $G$. Then the degree of each pair in this new bipartite graph $G^*$ (on vertex set $X cup Y^*$) is at most $1$ (since there is no $C_4$ in $G$) and the degree of each $xin X$ is at least $displaystyle3nover 2choose 2$. So we have $$kcdot 3nover 2choose 2leq lchoose 2 .$$
Since we assume $l<k$ we have $$3nover 2choose 2< k-1over 2$$ so $$3n(3n-2)over 4 < k-1 .$$ Since $kleq n^2$ we have $$3n(3n-2) leq 4n^2-8$$so
$$5n^2leq 6n-8$$ which is obviously not true.

Edit: After Darij's confirmation that the proof is correct, I'm will award any solution with better bound than $delta (G)=nsqrt2$ (instead of $3n/2$).

combinatorics discrete-mathematics proof-verification graph-theory

share|cite|improve this question

edited 12 hours ago

Maria Mazur

asked May 25 '18 at 15:10

Maria MazurMaria Mazur

46.7k1260120

$endgroup$

This question has an open bounty worth +100
reputation from Maria Mazur ending ending at 2019-03-17 18:50:50Z">in 4 days.

Looking for an answer drawing from credible and/or official sources.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  How exactly do you prove this for $n$ instead of $3n/2$ ? I don't get the inequalities to product a contradiction.
  $endgroup$
  – darij grinberg
  2 days ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  We are talking now about $delta = 3n/2$ or $n$?
  $endgroup$
  – Maria Mazur
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I am talking about $n$.
  $endgroup$
  – darij grinberg
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes, I see. My bad. Anyway, is the rest correct?
  $endgroup$
  – Maria Mazur
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The rest is correct. You may want to compute the optimal bound for your proof to work; I'm pretty sure it falls closer to $n$ than to $3n/2$.
  $endgroup$
  – darij grinberg
  2 days ago
  

  
  
  
  

 | 
show 2 more comments

$begingroup$

Say $G$ is a bipartite graph with bipartition $(A,B)$ and $G$ is $C_4$-free. Prove that if every vertex in $A$ has degree at least $frac32 n$ and $|A|leq n^2$, then $G$ has a matching which uses every vertex in $A$.

---

My proof:

Use the Hall marriage theorem. Take any $Xsubseteq A$ and let $|X|=k$. Also let $Y=N(X)$ and $|Y|=l$. Let us prove that $lgeq k$. Assume that $l<k$.

Let $Y^*$ be a set of all unordered pairs $y_i,y_j$, $ine j$ of elements in $Y$, and connect a pair $y_i,y_j$ with $xin X$ iff both $y_i$ and $y_j$ are adjacent with $x$ in $G$. Then the degree of each pair in this new bipartite graph $G^*$ (on vertex set $X cup Y^*$) is at most $1$ (since there is no $C_4$ in $G$) and the degree of each $xin X$ is at least $displaystyle3nover 2choose 2$. So we have $$kcdot 3nover 2choose 2leq lchoose 2 .$$
Since we assume $l<k$ we have $$3nover 2choose 2< k-1over 2$$ so $$3n(3n-2)over 4 < k-1 .$$ Since $kleq n^2$ we have $$3n(3n-2) leq 4n^2-8$$so
$$5n^2leq 6n-8$$ which is obviously not true.

Edit: After Darij's confirmation that the proof is correct, I'm will award any solution with better bound than $delta (G)=nsqrt2$ (instead of $3n/2$).

combinatorics discrete-mathematics proof-verification graph-theory

share|cite|improve this question

edited 12 hours ago

Maria Mazur

asked May 25 '18 at 15:10

Maria MazurMaria Mazur

46.7k1260120

$endgroup$

share|cite|improve this question

edited 12 hours ago

Maria Mazur

asked May 25 '18 at 15:10

Maria MazurMaria Mazur

46.7k1260120

share|cite|improve this question

edited 12 hours ago

Maria Mazur

asked May 25 '18 at 15:10

Maria MazurMaria Mazur

46.7k1260120

asked May 25 '18 at 15:10

Maria MazurMaria Mazur

46.7k1260120

asked May 25 '18 at 15:10

Maria MazurMaria Mazur

46.7k1260120